It follows that the general solution of the system is $x_1=i x_2$.
Hence, we have
\[E_{a+ib} =\Span \left(\, \begin{bmatrix}
i \\
1
\end{bmatrix} \,\right).\]

Note that the other eigenvalue $a-ib$ is the complex conjugate of $a+ib$.
It follows that the eigenspace $E_{a-ib}$ is obtained by conjugating the eigenspace $E_{a+ib}$.
Thus,
\[E_{a-ib} =\Span \left(\, \begin{bmatrix}
-i \\
1
\end{bmatrix} \,\right).\]

(c) Diagonalize the matrix $A$

From part (b), we see that
\[\begin{bmatrix}
i \\
1
\end{bmatrix} \text{ and } \begin{bmatrix}
-i \\
1
\end{bmatrix}\]
form an eigenbasis for $\C^2$.

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